Problem 16 - Precalculus Mathematics for Calculus

Question

Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. $$x^{2}-y^{2}+4=0$$

Step-by-Step Solution

Answer

Vertices: (0, ±2); Foci: (0, ±2√2); Asymptotes: y = ±x.

Step 1: Rewrite the Equation

Start with the given equation: \(x^2 - y^2 + 4 = 0\). Move the constant to the right side to get \(x^2 - y^2 = -4\). Since standard hyperbola equations are of the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), divide every term by \(-4\) to obtain \(-\frac{x^2}{4} + \frac{y^2}{4} = 1\). After rearranging, the equation of the hyperbola becomes \(\frac{y^2}{4} - \frac{x^2}{4} = 1\).

Step 2: Identify Hyperbola Type and Components

From \(\frac{y^2}{4} - \frac{x^2}{4} = 1\), recognize this as a vertically oriented hyperbola (since \(y^2\) term is positive), with \(a^2 = 4\) and \(b^2 = 4\), so \(a = 2\) and \(b = 2\).

Step 3: Find the Vertices

For a vertically oriented hyperbola centered at the origin, vertices are at \((0, \pm a)\). Therefore, the vertices of this hyperbola are \((0, \pm 2)\).

Step 4: Determine the Foci

The distance to each focus from the center is \(c\), where \(c^2 = a^2 + b^2\). Since \(a^2 = 4\) and \(b^2 = 4\), then \(c^2 = 8\), resulting in \(c = \sqrt{8} = 2\sqrt{2}\). Thus, the foci are at \((0, \pm 2\sqrt{2})\).

Step 5: Find the Asymptotes

For a hyperbola of the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the equations of the asymptotes are \(y = \pm\frac{a}{b}x\). Substituting \(a = 2\) and \(b = 2\), the asymptotes are \(y = \pm x\).

Step 6: Sketch the Hyperbola

To sketch the hyperbola, draw a rectangle centered at the origin with vertices at \((\pm b, \pm a)\), which are \((\pm 2, \pm 2)\). Draw the asymptotes as diagonal lines through the origin (lines with slopes \(\pm 1\)). The hyperbola branches will approach but never touch these lines, passing through the vertices \((0, \pm 2)\).

Key Concepts

Vertices of Hyperbola

Let's explore the vertices of a hyperbola, which are key points that mark the tips of the hyperbola along its major axis. In the standard form of a hyperbola, these vertices are found based on the term that is positive.

For our problem, the hyperbola's equation \[ \frac{y^2}{4} - \frac{x^2}{4} = 1 \]indicates a vertically oriented hyperbola since the \(y^2\) term is positive. This means the major axis is vertical, and the vertices are a distance \(a\) from the center along the \(y\)-axis.

Given that \(a^2 = 4\), we calculate \(a = 2\). Thus, the vertices are at the coordinates

• \((0, 2)\)
• \((0, -2)\)

These points are important as they help determine the shape and direction of the hyperbola.

Foci of Hyperbola

The foci of a hyperbola are points located on its major axis, farther from the center than the vertices. They play a critical role because hyperbolas curve around these foci.

The formula used to find the distance \(c\) to the foci from the center is\[ c^2 = a^2 + b^2 \]where \(a\) and \(b\) are determined from the hyperbola's standard form. Plugging the values for our problem, \(a^2 = 4\) and \(b^2 = 4\), we get:
\[ c^2 = 4 + 4 = 8 \quad \Rightarrow \quad c = \sqrt{8} = 2\sqrt{2} \]This shows us that the foci are at positions

• \((0, 2\sqrt{2})\)
• \((0, -2\sqrt{2})\)

These positions help us further understand how the hyperbola stretches and centers around the foci.

Asymptotes of Hyperbola

Asymptotes are crucial for understanding the behavior of a hyperbola. They are lines that the branches of a hyperbola approach but never intersect.

For our vertically oriented hyperbola with the equation \[ \frac{y^2}{4} - \frac{x^2}{4} = 1 \],the asymptotes have the equations \[ y = \pm \frac{a}{b} x \].Given \(a = 2\) and \(b = 2\), the asymptotes simplify to:
\[ y = \pm x \]These are straight lines through the origin with slopes \(+1\) and \(-1\). They guide the shape of the hyperbola by determining the direction of its branches.

• Asymptote 1: \( y = x \)
• Asymptote 2: \( y = -x \)

When sketching the hyperbola, drawing these asymptotes first helps visualize how the hyperbola bends towards these lines while forming its signature swooping curves.